Why signs decide this chapter
Take a reaction with ΔH = −100 kJ and ΔS = −200 J/K at 300 K. The calculation is one line, and the answer depends entirely on handling two negatives correctly: ΔG = −100 − (300)(−0.200) = −100 + 60 = −40 kJ. Getting the second sign wrong gives −160 kJ, which is not merely inaccurate but changes the reasoning about what happens at higher temperatures.
The same applies to work, to bond enthalpies and to the ΔH-versus-ΔU conversion. There is very little to understand and a great deal to keep straight, which makes this a chapter where writing out each step pays for itself.
How JEE actually asks Chemical Thermodynamics
NTA publishes no chapter-wise weightage, so any figure circulating online is a coaching estimate from past papers; confirm your session's syllabus and pattern in the official NTA information bulletin.
The recurring types are: apply the first law to a described process, convert between ΔH and ΔU, use Hess's law or bond enthalpies to find a reaction enthalpy, and determine spontaneity from ΔH and ΔS. Conceptual questions on state functions and the second law appear alongside them and are quick marks when the definitions are secure.
Key concepts, compressed
- ✓A state function depends only on the current state, not the path taken - internal energy, enthalpy, entropy and free energy are all state functions.
- ✓Work and heat are path functions: their values depend on how the change was carried out.
- ✓The first law is conservation of energy applied to a thermodynamic system.
- ✓Enthalpy is defined as H = U + PV, which makes ΔH equal to the heat exchanged at constant pressure.
- ✓Entropy measures the dispersal of energy, and the second law requires the total entropy of the universe to increase in any spontaneous process.
- ✓Gibbs free energy combines enthalpy and entropy into a single criterion for spontaneity at constant temperature and pressure.
Relations you need before attempting the questions
| Quantity | Formula | Note |
|---|---|---|
| First law | ΔU = q + w | work on the system is positive |
| Work of expansion | w = −P(ext)ΔV | against constant external pressure |
| Enthalpy | H = U + PV | |
| ΔH and ΔU | ΔH = ΔU + Δn(g)RT | gas moles only |
| Hess's law | ΔH is path-independent | enthalpy is a state function |
| From bond enthalpies | ΔH = Σ(bonds broken) − Σ(bonds formed) | order matters |
| Entropy change | ΔS = q(rev)/T | |
| Second law | ΔS(universe) > 0 for a spontaneous process | |
| Gibbs free energy | ΔG = ΔH − TΔS | T in kelvin |
| Spontaneity | ΔG < 0 | at constant T and P |
| Free energy and K | ΔG° = −RT ln K | |
| Transition temperature | T = ΔH/ΔS | where ΔG = 0 |
The five mistakes that cost the most marks
- ✓Getting the sign of work wrong. With ΔU = q + w, expansion by the system gives negative work; different textbooks use ΔU = q − w, so fix one form and stay with it.
- ✓Reversing the bond-enthalpy subtraction. Breaking bonds absorbs energy and forming them releases it, so ΔH is broken minus formed.
- ✓Using ΔH where ΔU is required. They differ by Δn(g)RT and coincide only when the gas-mole count is unchanged.
- ✓Treating spontaneous as meaning fast. Diamond converting to graphite is spontaneous and takes geological time.
- ✓Forgetting to convert ΔS from joules to kilojoules before combining it with ΔH. This factor of 1000 is the chapter's most common arithmetic failure.
Practice set 1: systems and the first law
1. What are open, closed and isolated systems?
An open system exchanges both matter and energy with its surroundings; a closed system exchanges energy only; an isolated system exchanges neither. A cup of coffee is open, a sealed flask is closed, and an ideal thermos approximates isolated. The distinction determines which quantities can change.
2. What is the difference between a state function and a path function?
A state function depends only on the initial and final states - internal energy, enthalpy, entropy and free energy are examples. A path function depends on how the change was achieved, and heat and work are the two that matter here. This is precisely why Hess's law works: ΔH is a state function, so the route is irrelevant.
3. State the first law of thermodynamics.
ΔU = q + w: the change in internal energy equals the heat added to the system plus the work done on it. It is conservation of energy applied to a thermodynamic system - energy is neither created nor destroyed, only transferred as heat or work.
4. In the convention ΔU = q + w, what signs do heat absorbed and work of expansion carry?
Heat absorbed by the system is positive; work done by the system in expanding is negative. The reasoning is that expanding against an external pressure costs the system energy. Some textbooks write ΔU = q − w with w defined as work done by the system, which gives identical physics with reversed bookkeeping - never mix the two.
5. A gas expands from 1 L to 3 L against a constant external pressure of 2 atm. How much work is done on the gas?
About −405 J. Work is w = −P(ext)ΔV = −(2 atm)(2 L) = −4 L·atm. Converting with 1 L·atm = 101.3 J gives −405.2 J. The negative sign means the system does work on the surroundings, losing energy - and if no heat were supplied, its internal energy would fall by exactly that amount.
6. Name the four standard processes and what is held constant in each.
Isothermal holds temperature constant, adiabatic exchanges no heat with the surroundings, isochoric holds volume constant, and isobaric holds pressure constant. Each simplifies the first law differently - for instance an adiabatic process has q = 0, so ΔU = w, and an isochoric process has no expansion work.
7. What does the heat absorbed at constant volume equal?
The change in internal energy, ΔU, because no expansion work is possible when the volume cannot change. At constant pressure the heat absorbed equals ΔH instead. This is the practical distinction between the two quantities, and it is why bomb calorimetry measures ΔU while ordinary calorimetry measures ΔH.
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Practice set 2: enthalpy
8. How is enthalpy defined and why is it useful?
H = U + PV. Its usefulness is that ΔH equals the heat exchanged at constant pressure, which is the condition most laboratory reactions occur under. Since we usually work in open vessels at atmospheric pressure, ΔH is the quantity that a thermometer effectively measures.
9. How are ΔH and ΔU related for a reaction involving gases?
ΔH = ΔU + Δn(g)RT, where Δn(g) is the change in the number of moles of gas. The difference arises because a change in gas moles causes expansion or contraction work at constant pressure. Only gaseous species count - solids and liquids contribute negligibly.
10. For a reaction with Δn(g) = −2 at 300 K, what is the difference between ΔH and ΔU?
About −4.99 kJ/mol. The difference is Δn(g)RT = (−2)(8.314)(300) = −4988.4 J, roughly −4.99 kJ. So ΔH is about 5 kJ more negative than ΔU. When gas moles decrease, the surroundings do work on the system, which is what the negative sign records.
11. What do exothermic and endothermic mean in terms of ΔH?
Exothermic means ΔH is negative - the system releases heat and the surroundings warm. Endothermic means ΔH is positive - the system absorbs heat. The sign is from the system's perspective throughout, which is worth stating explicitly because the surroundings experience the opposite.
12. State Hess's law and explain why it works.
The enthalpy change of a reaction is the same whether it occurs in one step or several. It works because enthalpy is a state function, so ΔH depends only on the initial and final states, not the route. Practically, it lets you calculate an enthalpy change that cannot be measured directly by combining reactions that can.
13. How is a reaction enthalpy calculated from bond enthalpies?
ΔH = Σ(bond enthalpies of bonds broken) − Σ(bond enthalpies of bonds formed). Breaking bonds requires energy and forming them releases it, which is what fixes the order. Reversing the subtraction gives the correct magnitude with the wrong sign, turning an exothermic reaction into an endothermic one.
14. What is the standard enthalpy of formation of an element in its standard state?
Zero, by definition. Forming an element from itself involves no change. This convention is what makes standard enthalpies of formation usable: the enthalpy of a reaction is the sum for the products minus the sum for the reactants, and elements simply drop out of that calculation.
Practice set 3: entropy
15. What does entropy measure?
The dispersal of energy among the available states of a system, commonly described as a measure of disorder. A higher entropy means the energy is spread over more accessible arrangements. It is a state function with an absolute value, unlike enthalpy, which is only ever measured as a change.
16. What happens to entropy during melting and boiling?
It increases in both, because particles gain freedom of movement. The increase on boiling is much larger than on melting, since the jump from liquid to gas involves a far greater expansion of accessible states than the jump from solid to liquid. Freezing and condensing have negative ΔS for the system.
17. State the second law of thermodynamics.
The total entropy of the universe increases in any spontaneous process. Note that it applies to the universe - system plus surroundings - not to the system alone. That distinction is what allows a system's entropy to decrease spontaneously, provided the surroundings gain more.
18. State the third law of thermodynamics.
The entropy of a perfect crystalline substance is zero at absolute zero, because there is only one possible arrangement of the particles. This provides the reference point that makes absolute entropies measurable, which is why tables of S° exist while tables of absolute enthalpy do not.
19. How is entropy change calculated for a reversible process?
ΔS = q(rev)/T, the heat exchanged reversibly divided by the absolute temperature. Because entropy is a state function, this value applies even when the actual process is irreversible - you compute it along a reversible path between the same two states. The temperature must be in kelvin.
20. How do the entropies of solids, liquids and gases compare?
Gas is much greater than liquid, which is greater than solid. Particles in a gas have the most freedom of position and motion and therefore the most accessible arrangements. This ordering lets you predict the sign of ΔS for many reactions by simply counting gas moles on each side.
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Practice set 4: Gibbs free energy
21. Write the Gibbs free energy relation.
ΔG = ΔH − TΔS, with T in kelvin. It combines the enthalpy and entropy contributions into one criterion, valid at constant temperature and pressure. The two terms can oppose each other, which is why the outcome often depends on temperature.
22. What is the condition for spontaneity?
ΔG < 0. A negative free energy change means the process can occur without external input at constant temperature and pressure. ΔG = 0 means the system is at equilibrium, and ΔG > 0 means the reverse process is the spontaneous one.
23. Describe the four possible sign combinations of ΔH and ΔS.
Negative ΔH with positive ΔS is spontaneous at all temperatures. Positive ΔH with negative ΔS is never spontaneous. Both negative is spontaneous only at low temperature, since the −TΔS term becomes positive and grows with T. Both positive is spontaneous only at high temperature, where the favourable entropy term dominates.
24. What does ΔG = 0 signify?
Equilibrium - there is no net drive in either direction. It is also the condition used to find a transition temperature, such as a melting point, where two phases coexist. Setting ΔH − TΔS = 0 gives T = ΔH/ΔS, which is how question 28 is answered.
25. How is standard free energy related to the equilibrium constant?
ΔG° = −RT ln K. A negative ΔG° corresponds to K greater than 1, meaning the products are favoured at equilibrium. Note that ΔG° is a fixed value for the reaction while ΔG varies with composition and is zero at equilibrium - the two are distinct and questions exploit the confusion.
26. Does a negative ΔG mean the reaction will be fast?
No. ΔG describes thermodynamic feasibility only, and says nothing about rate. The conversion of diamond to graphite has a negative ΔG and proceeds imperceptibly slowly because the activation energy is enormous. Rate is the domain of kinetics, and the two are entirely independent.
Practice set 5: calculations
27. A reaction has ΔH = −100 kJ and ΔS = −200 J/K at 300 K. Is it spontaneous?
Yes - ΔG = −40 kJ. Convert ΔS to kJ first: −0.200 kJ/K. Then ΔG = ΔH − TΔS = −100 − (300)(−0.200) = −100 + 60 = −40 kJ. Note the double negative: subtracting a negative TΔS adds 60 kJ. The reaction is spontaneous, but only because the temperature is low enough.
28. Above what temperature does that reaction cease to be spontaneous?
500 K. Set ΔG = 0, giving T = ΔH/ΔS = (−100)/(−0.200) = 500 K. Both quantities are negative so the ratio is positive. Below 500 K the enthalpy term dominates and the reaction is spontaneous; above it the unfavourable entropy term takes over. This is the classic both-negative case from question 23.
29. How do you apply Hess's law numerically?
Arrange the given equations so that they add to the target equation, reversing any as needed - which flips the sign of its ΔH - and multiplying where necessary, which scales ΔH by the same factor. Then add the adjusted enthalpy values. The two operations to watch are the sign flip on reversal and scaling ΔH along with the coefficients.
30. What is the most common unit error in this chapter?
Mixing kilojoules and joules when combining ΔH with TΔS. Enthalpies are usually tabulated in kJ/mol and entropies in J/mol·K, so one must be converted before they are combined. Failing to do so produces an answer wrong by a factor of a thousand, which usually reverses the conclusion about spontaneity.
How to study this chapter efficiently
- ✓Write the sign convention at the top of your working and use it consistently. Do not switch between ΔU = q + w and ΔU = q − w.
- ✓Convert ΔS to kilojoules before combining it with ΔH, every time. Make it a reflex rather than a check.
- ✓For bond enthalpies, write 'broken minus formed' explicitly before substituting numbers.
- ✓Memorise the four sign combinations of ΔH and ΔS as a small table. They answer most spontaneity questions instantly.
- ✓Keep ΔG and ΔG° distinct: the standard value fixes K, the actual value is zero at equilibrium.
- ✓Prepare this chapter before Electrochemistry, which uses the ΔG relation directly.
Turn this into active practice
Because the errors here are almost all mechanical, they are also invisible when reading. A worked solution shows the double negative already resolved; only producing one yourself reveals whether you handle it reliably.
The JEE Thermodynamics quiz on QUFF generates fresh questions across the first law, enthalpy, entropy and free energy, marks them instantly and explains each answer. Do a timed set and classify each error as a sign error, a unit error or a conceptual one. The first two will dominate, and both are entirely fixable.
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